Usual speed 2

A passenger train takes 1 h less for a journey of 120 km.If its speed increased by 10 k/h from its usual speed. What is its usual speed?

Correct answer:

v =  30 km/h

Step-by-step explanation:

Δv=10 km/h Δt=1 h s=120 km  s = v t = (v+Δv) (tΔt)  v t = (v+Δv) (tΔt) v t = v t+Δv   tΔt vΔt Δv 0 = Δv   tΔt vΔt Δv Δt v+Δt Δv = Δv   t Δt v+Δt Δv = Δv   (s/v)  Δt v2+Δt Δv v=Δv s  1 v2+1 10 v=10 120 v2+10v1200=0  a=1;b=10;c=1200 D=b24ac=10241(1200)=4900 D>0  v1,2=2ab±D=210±4900 v1,2=210±70 v1,2=5±35 v1=30 v2=40  v>0  v=v1=30 km/h   Verifying Solution:  t1=s/v=120/30=4 h t2=v+Δvs=30+10120=3 h

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